Prove that; (1+cotA+TanA).(SinA-CosA)=Sec3A-Cosec3A/Sec2A.Cosec2A.
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Answer:
LHS
=
sec
3
A−cosec
3
A
(1+cotA+tanA)(sinA−cosA)
=
(secA−cosecA)(sec
2
A+cosec
2
A+secAcosecA)
(1+cotA+tanA)(sinA−cosA)
=
(sec
2
A+cosec2A+secAcosecA)
(1+cotA+tanA)(sinA−cosA)
=
(cosAsinA+cos
2
A+sin
2
A)
(1+cotA+tanA)(sin
3
Acos
3
A)
=
(sinAcosA+1)
(1+
sinA
cosA
+
cosA
sinA
)(sin
3
Acos
3
A)
=
(1+sinAcosA)
(1+sinAcosA)(sin
2
Acos
2
A)
=sin
2
Acos
2
A=RHS
Hence proved
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